题目链接: B. Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mary has just graduated from one well-known University and is now attending celebration party. Students like to d…
B. Pyramid of Glasses 题目连接: http://www.codeforces.com/contest/676/problem/B Description Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne…
B. Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream o…
原题链接 B. Pyramid of Glasses Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyram…
Pyramid of Glasses time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a…
题目链接:http://codeforces.com/problemset/problem/676/B 题意:一个n层的平面酒杯金字塔,如图,每个杯子的容量相同.现在往最顶部的一个杯子倒 t 杯酒,求流动结束后有多少个杯子被装满. 思路:每个局部的两代三个杯子的流动过程是一致的,因此可以用递归来模拟求解. 具体为:设add(cur, i, l)执行“往第 i 个杯子倒cur量的酒”,附加信息: i 位于第 l 层.若执行完剩余了surplus量的酒,则往 i 的下一层左右两侧的杯子各倒surpl…
题意:每秒从最高处的杯子倒一杯酒下来,酒流的方式如图,问t秒装满酒的杯子的数目. 思路:把第一杯的值设为t,glass[i][j]=(glass[i-1][j-1]-1)/2+(glass[i-1][j])/2,统计值大于等于1的杯数. #include<cstdio> ][]; int main() { int n,t; scanf("%d%d",&n,&t); glass[][]=t; ;i<=n;i++) { ;j<=n;j++) { ][…
题目链接:http://codeforces.com/problemset/problem/723/B 题目大意: 输入n,给出n个字符的字符串,字符串由 英文字母(大小写都包括). 下划线'_' .括号'(' ')' 组成.[括号不会嵌套] 求括号外面的连续字符串最大的字符串长度和括号内的连续字符串的个数. 举例: input 37_Hello_Vasya(and_Petya)__bye_(and_OK) output 5 4 括号外面的连续的字符串最长的字符串长度为 5 括号内有 4 个连续…
题目链接: http://codeforces.com/problemset/problem/546/C 题意: 总共有n张牌,1手中有k1张分别为:x1, x2, x3, ..xk1,2手中有k2张,分别为:y1, y2, ...yk2:(n<=10&&k1+k2==n,所有牌的数字都不同): 依次比较x1, y1的大小,若x1>y1,依次将x1, y1加入x牌最底下:反之则将y1,x1依次加入y牌最底下:直至有方的牌输完为止:输出总共游戏的步数和赢方: 如果两方都不能赢,则…
http://codeforces.com/problemset/problem/749/C 题意:有n个人投票,分为 D 和 R 两派,从1~n的顺序投票,轮到某人投票的时候,他可以将对方的一个人KO,被KO的人不能投票了,这样循环,直到某一派的人全部被KO,问哪方赢. 思路:模拟..代码好懂.. #include <cstdio> #include <algorithm> #include <iostream> #include <cstring> #i…